each piece of the function is a polynomial, so we know each piece is continuous on its own. however, we must be concerned with where the function breaks. this is at x = 3.
for 1a, k = 1 does not work because when we plug in k = 1 and x = 3 we end up with two different values for each piece, which obviously (why is this obvious?) makes the function not continuous.
for 1b. we need to find k so that the function is continuous. meaning, when x = 3, we want both functions to give the same value. how do you think we should proceed?
*Ahem* b00yeah05 ........ which is why I wrote -(3)^2, NOT (-3)^2:Originally Posted by b00yeah05
A flagon of frustration might have been saved if the original equation (above) that started this almost-death by a thousand cuts had been read and worked a little more ...... carefully?
No need to thank me for the above (apparently useless) post - the pleasure's in the giving.